ME101 Lecture09 11 HBK

ME101 - Division I L09-L11

Hemant B Kaushik 1

2012-2013 (II Semester)Division I

Instructor

ME101: Engineering Mechanics (3 1 0 8)

Instructor

Hemant B. Kaushik

Room M-003Department of Civil Engineering

IIT Guwahati

ME101 - Division I Hemant B Kaushik 1

[email protected]: 258 2427

Web: www.iitg.ernet.in/hemantbkshilloi.iitg.ernet.in/~hemantbk/ME101.htm

L09

Beams Internal Effects Internal Force Resultants Axial Force (N), Shear Force (V), Bending Moment

(M), Torsional Moment (T) in Beam( ), ( ) Method of Sections is used

ME101 - Division I Hemant B Kaushik 2Axial

Beams Internal Effects Method of Section:

Internal Force Resultants at B Section a-a at B and use equilibrium equations in both cut parts


Beams Internal Effects 2D Beam

3D Beam


The Force Resultants act atcentroid of the sections Cross-sectional area


Hemant B Kaushik 2

Beams Internal EffectsSign ConventionPositive Axial Force creates Tension

Positive shear force will cause the Beam segment on which it acts to rotate clockwise

Positive bending moment will tend to bend the segment on which it acts in a concave upward manner (compression on top of section).


Beams Internal EffectsSign convention in a single plane

H-section Beam bent by two equal and opposite positive moments applied at ends

Neglecting resistance offered by web

Interpretation of Bending Couple


Compression at top; Tension at bottom

Resultant of these two forces (one tensile and other compressive) acting on any section is a Couple and has the value of the Bending Moment acting on the section.

Beams Internal EffectsExample: Find the axial force in the fixed bar at points B and CSolution: Draw the FBD of the entire bar

Draw sections at B and C to get AF in the bar at B and C


Alternatively, take a section at C and consider only CD portion of the barThen take a section at B and consider only BD portion of the bar no need to calculate reactions

Beams Internal EffectsExample: Find the internal torques at points B and C of the circular shaft subjected to three concentrated torquesSolution: FBD of entire shaft

Sections at B and C and FBDs of shaft segments AB and CD


Sections at B and C and FBDs of shaft segments AB and CD


Hemant B Kaushik 3

Beams Internal EffectsExample: Find the AF, SF, and BM at point B (just to the left of 6 kN) and at point C (just to the right of 6 kN) Solution: Draw FBD of entire beam


Dy need not be determined if only left part of the beam is analysed

Beams Internal EffectsExample Solution: Draw FBD of segments AB and ACand use equilibrium equations


Beams SFD and BMDShear Force Diagram and bending Moment Diagram Variation of SF and BM over the length of the beam SFD and BMD

Maximum magnitude of BM and SF and their locations is prime consideration in beam design

SFDs and BMDs are plotted using method of section Equilibrium of FBD of entire Beam External Reactions External Reactions

Equilibrium of a cut part of beam Expressions for SF and BM at the cut section

Use the positive sign convention consistently

ME101 - Division I Hemant B Kaushik 11L10

Beams SFD and BMDDraw SFD and BMD for a cantilever beam supporting a point load at the free end

F

L/4 L/4 L/4 L/4

+V

-V

+F SFD+F+F+F+F+F


+M

-MBMD

-FL-3FL/4

-FL/2-FL/4

M=0


Hemant B Kaushik 4

Beams SFD and BMDShear and Moment RelationshipsConsider a portion of a beam Isolate an element dxD FBD f th l tDraw FBD of the element

Vertical equilibrium in dx

dxdVw =V w dx (V + dV) = 0

Moment equilibrium in dx (M@ the left side of the element)

Terms w(dx)2/2 and dVdx may be dropped since they are differentials of higher order than those which remains.


( ) 0)(2

=++++ dMMdxdVVdxwdxMdxdMV =

Beams SFD and BMDShear and Moment Relationships

dVw =Slope of the shear diagram = - Value of applied loading

dxw

dxdMV =

Slope of the moment curve = Shear Force

Both equations not applicable at the point of loading because of discontinuity produced by the abrupt change in shear.


Beams SFD and BMDShear and Moment RelationshipsExpressing V in terms of w by integrating

xVdxdVw =

OR

V0 is the shear force at x0 and V is the shear force at x

Expressing M in terms of V by integrating

V = V0 + (the negative of the area underthe loading curve from x0 to x)

= xxVV wdxdV 00

dxdMV =

OR

M0 is the BM at x0 and M is the BM at x


= xxMM VdxdM 00 M = M0 + (area under the shear diagram from x0 to x)

Beams SFD and BMDV = V0 + (negative of area under the loading curve from x0 to x)

M = M0 + (area under the shear diagram from x0 to x)

If there is no externally applied moment M0 at x0 = 0,total moment at any section equals the area underthe shear diagram up to that section

When V passes through zero and is a continuousfunction of x with dV/dx 0 (i.e., nonzero loading) 0=dM

BM will be a maximum or minimum at this point

Critical values of BM also occur when SF crosses the zero axis discontinuously (Example: Beams under concentrated loads)


dx


Hemant B Kaushik 5

Beams SFD and BMDDegree of V in x is one higher than that of w

Degree of M in x is one higher than that of V

dxdVw =

dxdMV =

Degree of M in x is two higher than that of w

Combining the two equations

If i k f ti f BM b bt i d b

dx

wdx

Md =22

If w is a known function of x, BM can be obtained by integrating this equation twice with proper limits of integration.Method is usable only if w is a continuous function of x (other cases

not part of this course)ME101 - Division I Hemant B Kaushik 17

Beams SFD and BMD: Example Draw the SFD and BMD.

Determine reactions at Determine reactions at supports.

Cut beam at C and consider member AC,

22 PxMPV +=+= Cut beam at E and consider

member EB,

( ) 22 xLPMPV +== For a beam subjected to

concentrated loads, shear is constant between loading points and moment varies linearly.


Maximum BM occurs where Shear changes the direction

Beams SFD and BMD: ExampleDraw the shear and bending moment diagrams for the beam and loading shown.

Solution: Draw FBD and find out thesupport reactions using equilibrium equations

ME101 - Division I Hemant B Kaushik 19L11

Beams SFD and BMD: Example

= :0yF 0kN20 1 = V kN201 =V:01 =M ( )( ) 0m0kN20 1 =+ M 01 =M

Use equilibrium conditions at all sections toget the unknown SF and BM

( )( ) 00N0 1 01V2 = -20 kN; M2 = -20x kNmV3 = +26 kN; M3 = -20x+460 = -20x MB= -50 kNmV4 = +26 kN; M4 = -20x+46(x-2.5) MC= +28 kNmV5 = -14 kN; M5 = -20x+46(x-2.5)-400 MC= +28 kNmV6 = -14 kN; M6 = -20x+46(x-2.5)-40(x-5.5)

MD= 0 kNm


Alternatively, from RHSV5 = V6 = -14 kNM5 = +14x MC= +28 kNmM6 = +140 MD= 0 kNmImportant: BM0 at Hinged Support B. Why?


Hemant B Kaushik 6

Beams SFD and BMD: ExampleDraw the SFD and BMD for the beam acted upon by a clockwise couple at mid point

CSolution: Draw FBD of the beam and Calculate the support reactions

Draw the SFD and the BMD starting

ClC

lC

V

lC


gFrom any one end

l

M

2C

2C

Beams SFD and BMD: ExampleDraw the SFD and BMD for the beam

Solution: Draw FBD of the beam and Calculate the support reactionsCalculate the support reactions

Draw the SFD and the BMD starting

MA = 0 RA = 60 N MB = 0 RB = 60 N

120 Nm60 N

60 NV

-60 N


gfrom any one end

60 N

M

-120 Nm

Beams SFD and BMD: ExampleDraw the SFD and BMD for the beam

Solution:Draw FBD of the entire beamDraw FBD of the entire beamand calculate support reactionsusing equilibrium equations

Reactions at supports:2

wLRR BA ==

w


Develop the relations between loading, shear force, and bending moment and plot the SFD and BMD

Beams SFD and BMD: Example

Shear Force at any section:

== wxdxwVVx

A

w

== xLwwxwLV22

Alternatively,

=== xLwwxwLwxVV A 22

0

BM at any section:

0

VdxMMx

A = ( )2

222xxLwxwxxwLM ==

Alternatively,


( )20

0

22xxLwdxxLwM

x

=

=

=== 0at 8

2

max VdxdMMwLM Q


Hemant B Kaushik 7

Beams SFD and BMD: ExampleDraw the SFD and BMD for the Beam

Solution:SFD and BMD can be plotted withoutdetermining support reactions sinceit is a cantilever beam

Area under SFD

it is a cantilever beam.However, values of SF and BM can be verified at the support if supportreactions are known.


( )aLawaLawMawR CC =

== 3632

;2

000

M

Beams SFD and BMD: Summary

2.5 m 3 m 2m

SFD

l/2 l/2

2 m 2 m

w

L

BMD

SFD

ME101 - Division I 26Hemant B Kaushik

BMD

ME101 Lecture09 11 HBK

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